Advertisements
Advertisements
Question
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Options
3
6
9
1
Advertisements
Solution
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to 9.
Explanation:
Since x = `int_0^y "dt"/sqrt(1 + 9"t"^2)`
⇒ `"dx"/"dy" = 1/sqrt(1 + 9y^2)`
which gives `("d"^2y)/("dx"^2) = (18y)/(2sqrt(1 + 9y^2)) * "dy"/"dx"`
= 9y.
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^4 x dx\]
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
